« first day (1065 days earlier)      last day (2363 days later) » 

BAYMAX
BAYMAX
3:03 AM
.
C(X,Y) is tehset of all compact operators from $X$ to $Y$.
$B(X,Y)$ is the set of all bounded linear operators from $X$ to $Y$
.
If $T \in C(X,Y)$ then $T \in B(X,Y)$ , how easily one sees this?
.
Like i thought like this, from the definition of $T \in C(X,Y)$ we see that for any bounded subset $B$ of $X$ , $\overline{T(B)} $ is bounded in $Y$.
then if the closure is bounded in $Y$ what can we say next?
perhaps then we can also say that $T(B)$ is also bounded
as $B$ is arbitrary bounded set
so $T$ is bounded
implying $T$ is continuous
 
BAYMAX
BAYMAX
3:25 AM
.
 
 
4 hours later…
BAYMAX
BAYMAX
7:18 AM
why in this following question answer here, $0 \leq l(1-f)$
?
 
BAYMAX
BAYMAX
7:31 AM
2) Let $B$ be a Banach Space (not finite dimensional) and $T : B \tightarrow B$ be a continuous operator such that range of $T$ is $B$ and $T(x) = 0$ imply $x = 0$,then
$T$ maps compact sets to opensets
also why it cannot map bounded sets to compact sets?
 

« first day (1065 days earlier)      last day (2363 days later) »